How can the friction drag be calculated for a moving and spinning disk on a 2D surface?

Question

Let's consider a disk with mass m and radius R on a surface where friction u also involved. When we give this disk a starting velocity v in a direction, the disk will go towards that direction and slow down and stop.

In case the disk has a rotation (or spin with the rotational line perpendicular on the surface) w beside the speed then the disk won't move on a line, instead bend. Both the linear and angular velocity would be 0 at the end.

How can this banding/curving/dragging be calculated? Is it possible to give analytical solution for the X(v,w,t) function, where X would give the position of the disk according to it's initial v w at a given t?

Any simulation hint would be also fine. I imagine, depending on w and m and u there would be an additional velocity which is perpendicular to the linear velocity and so the disk's path would bend from the linear path.

Answer 1

If you're going to simulate this, I'd probably recommend something like dividing up the contact surface between the disk and the table into a radial grid. Compute the relative velocity and the force at each point on the grid at each time step, then sum up the forces and torques (r cross F) to get the net force F and the net torque T on the disk as a whole. Then you can apply the equations F=(m)(dv/dt) and T=(I)(dw/dt) to determine the differential changes in v and w for the next time step.

For what it's worth, I don't think a flat disk would curve under the influence of either a frictional force (velocity-independent) or a drag force (linearly proportional to velocity).

Answer 2

A ball will move in a large arc with spin, but a [uniform] disk on a 2D surface will not.

For the disk it's center of spin is the same as it's center of gravity, so there is no torque applied. (As mentioned by duffymo, a nonuniform disk will have a torque applied.)

For a uniform ball, if the axis of the spin is not perpendicular to the table, this causes the ball to experience a rotational torque which causes it to move in a slight arc. The arc has a large radius, and the torque is slight, so usually friction makes the ball stop quickly.

If there was a sideways velocity, the ball would move along a parabola, like a falling object. The torque component (and the radius of the arc) can be computed in the same way you do for a precessing top. It's just that the ball sits at the tip of the top (err....) and the bottom is "imaginary".

Top equation: http://hyperphysics.phy-astr.gsu.edu/HBASE/top.html

omega_p = mgr/I/omega

where

omega_p = rotational velocity...dependent on how quickly you want friction to slow the ball
m = ball mass
g = 9.8 m/s^2 (constant)
r = distance from c.g. (center of ball) to center, depends on angle of spin axis (solve for this)
omega = spin rate of ball
I = rotational inertia of a sphere

My 2 cents.

Answer 3

Numerical integration of Newton's laws of motion would be what I'd recommend. Draw the free body diagram of the disk, give the initial conditions for the system, and numerically integrate the equations for acceleration and velocity forward in time. You have three degrees of freedom: x, y translation in the plane and the rotation perpendicular to the plane. So you'll have six simultaneous ODEs to solve: rates of change of linear and angular velocities, rates of change for two positions, and the rate of change of angular rotation.

Be warned: friction and contact make that boundary condition between the disk and the table non-linear. It's not a trivial problem.

There could be some simplifications by treating the disk as a point mass. I'd recommend looking at Kane's Dynamics for a good understanding of the physics and how to best formulate the problem.

I'm wondering if the bending of the path that you're imagining would occur with a perfectly balanced disk. I haven't worked it out, so I'm not certain. But if you took a perfectly balanced disk and spun it about its center there'd be no translation without an imbalance, because there's no net force to cause it to translate. Adding in an initial velocity in a given direction wouldn't change that.

But it's easy to see a force that would cause the disk to deviate from the straight path if there was an imbalance in the disk. If I'm correct, you'll have to add an imbalance to your disk to see bending from a straight line. Perhaps someone who's a better physicist than me could weigh in.